 # For the equation, write the value or values of the variable that make a denominator zero. These are the restrictions on the variable. Keeping the restrictions in mind, solve the equation. $$\dfrac{4}{x+5}+\dfrac{2}{x-5}=\dfrac{32}{x^2-25}$$

This question aims to find the solution to the given equation by taking into consideration the restrictions on the given function.

The fraction of two polynomials is said to be a rational expression. Such expression can be expressed as $\dfrac{a}{b}$ in which $a$ and $b$ both are polynomials. The product, sum, division, and subtraction of a rational expression can be carried out similarly as they are carried out for the polynomials. Rational expressions possess a good property that the application of arithmetic operations results in a rational expression as well. More generally, it is simple to find out the product or quotient of two or more rational expressions, but tricky to subtract or add as compared to the polynomials.

A function is said to be rational if there is at least one variable in the denominator of the rational expression. Let $h(y)$ and $k(y)$ be two functions in $y$ and $\dfrac{h(y)}{k(y)}$ be the rational function. A restriction on such a function can be defined as any value of the variable in the linear denominator that makes it zero. A restriction results in another function by selecting a relatively small domain for the rational function.

The restrictions on the domain can be found by equating the denominator to zero. The values of variables for which the denominator becomes zero and the function becomes undefined are said to be singularity and are excluded from the domain of the function.

## Numerical Results

For restrictions:

Let $x+5=0$, $x-5=0$ and $x^2-25=0$

$x=-5$, $x=5$ and $x=\pm 5$

So, the restrictions are $x=\pm 5$.

Now solve the given equation as:

$\dfrac{4}{x+5}+\dfrac{2}{x-5}=\dfrac{32}{x^2-25}$

$\dfrac{x-5}{x-5}\cdot\left(\dfrac{4}{x+5}\right)+\dfrac{x+5}{x+5}\cdot\left(\dfrac{2}{x-5}\right)=\dfrac{32}{x^2-25}$

$\dfrac{4(x-5)+2(x+5)}{(x-5)(x+5)}=\dfrac{32}{x^2-25}$

$\dfrac{4x-20+2x+10}{x^2-25}=\dfrac{32}{x^2-25}$

$\dfrac{6x-10}{x^2-25}=\dfrac{32}{x^2-25}$

$(x^2-25)\left(\dfrac{6x-10}{x^2-25}\right)=(x^2-25)\left(\dfrac{32}{x^2-25}\right)$

$6x-10=32$

$6x=32+10$

$6x=42$

$x=\dfrac{42}{6}$

$x=7$

## Example 1

Given below is a rational function with a non-linear denominator. Find the restrictions on the variable.

$\dfrac{2(x-2)}{x^2-4}$

### Solution

$\dfrac{2(x-2)}{x^2-4}=\dfrac{2(x-2)}{(x-2)(x+2)}$

$=\dfrac{2}{x+2}$

Now, to find the restrictions, equate the denominator to zero as:

$x+2=0$

$x=-2$

Since $x=-2$ makes the denominator zero and the given function undefined, this is the restriction on the variable.

## Example 2

Given below is a rational function with a linear denominator. Find the restrictions on the variable.

$\dfrac{3}{(3x-9)}$

### Solution

First, simplify the given expression as:

$\dfrac{3}{(3x-9)}=\dfrac{3}{3(x-3)}$

$=\dfrac{1}{x-3}$

Now, to find the restrictions, equate the denominator to zero as:

$x-3=0$

$x=3$

Since $x=3$ makes the denominator zero and the given function undefined, this is the restriction on the variable.